58 — Waves and Thermodynamics

Solving this equation, we get

A1 α +1
= Ans.
A2 α −1
2 2
I 1  A1   α + 1
(c) =   
I 2  A 2   α − 1

INTRODUCTORY EXERCISE 18.1

1. The ratio of intensities of two waves is 9:16. If these two waves interfere, then determine the
ratio of the maximum and minimum possible intensities.
2. In Interference two individual amplitudes are 5 units and 3 units. Find
Amax I max
(a) (b)
Amin I min
3. Three waves due to three coherent sources meet at one point. Their amplitudes are 2A0, 3A0
and 2A0. Intensity corresponding to A0 is I 0. Phase difference between first and second is 45°.
λ
Path difference between first and third . In phase angle, first wave lags behind from the other
4
two waves. Find resultant intensity at this point.

18.4 Standing Wave

Standing wave is an example of interference. When two identical waves travel in opposite directions,
then they superimpose almost at every point. Now, since the waves are identical (or their frequencies
are same), sources are coherent. So, interference will take place. By their superposition (or
interference) standing waves (also called stationary waves) are formed. By identical wave we mean
that all properties (like f , ω, T, λ and k) are same. Only amplitudes may be different but still we prefer
equal amplitudes ( A1 = A2 = A0 ). The following are listed some of the important points in standing
wave.
(i) Here, the constructive interference points are called antinodes (denoted by A) and destructive
interference points are called nodes (denoted by N ).
λ π  2π 
(ii) Distance between two successive nodes (or two successive antinodes) is or  as k =  .
2 k  λ
λ π
Similarly, distance between a node and its adjacent antinode is or . Here, λ and k are the
4 2k
values corresponding to constituent waves by which stationary waves are formed.
(iii) Amplitude of oscillation in stationary wave varies from a maximum value ( A1 + A2 or 2 A0 at
antinode) to a minimum value (A1 ~ A2 or zero at node). But now onwards (unless mentioned in
the question) we will talk about the equal amplitudes. So, amplitude at antinode will be 2 A0 and
at node it is zero.
Thus, now we can say that, in stationary wave all particles (except nodes for the case when
A1 = A2 ) oscillate with same frequency but different amplitudes.
 Chapter 18 Superposition of Waves — 63

(d) The equations of the component waves are

 πx 
y1 = 2sin  + 96 πt
 15 
 πx 
and y 2 = 2 sin  − 96 πt
 15 
as we can see that y = y1 + y 2

INTRODUCTORY EXERCISE 18.2

π
1. In stationary wave, phase difference between two particles can't be . Is this statement true or
3
false?
πx
2. A string vibrates according to the equation y = 5 sin cos 40 πt
3
where, x and y are in centimetres and t is in seconds
(a) What is the speed of the component wave?
(b) What is the distance between the adjacent nodes?
9
(c) What is the velocity of the particle of the string at the position x = 1.5 cm when t = s?
8
3. If two waves differ only in amplitude and are propagated in opposite directions through a
medium, will they produce standing waves. Is energy transported?
4. Two sinusoidal waves travelling in opposite directions interfere to produce a standing wave
described by the equation
y = (1.5 m )sin (0.400 x )cos (200 t )
where, x is in metres and t is in seconds. Determine the wavelength, frequency and speed of the
interfering waves.

18.5 Normal Modes of a String

In an unbounded continuous medium, there is no restriction on the frequencies or wavelengths of the
standing waves. However, if the waves are confined in space, for example, when a string is tied at
both ends,standing waves can be set-up for a discrete set of frequencies or wavelengths. Consider a
string of definite length l, rigidly held at both ends. When we setup a sinusoidal wave on such a string,
it gets reflected from the fixed ends. By the superposition of two identical waves travelling in
opposite directions standing waves are established on the string. The only requirement we have to
satisfy is that the end points be nodes as these points cannot oscillate. They are permanently at rest.
There may be any number of nodes in between or none at all, so that the wavelength associated with
the standing waves can take many different values. The distance between adjacent nodes is λ / 2, so
that in a string of length l there must be exactly an integral number n of half wavelengths λ /2. That is,
nλ
=l
2
2l
or λ= (n = 1, 2, 3 …)
n
 Chapter 18 Superposition of Waves — 67

In air, T = mg = (Vρ ) g
1 Vρg
∴ ν= …(i)
2l µ
When the object is half immersed in water, then
V  V 
T ′ = mg − upthrust = Vρg −   ρ w g =   g ( 2ρ − ρ w )
 2  2
The new fundamental frequency is
1 T′ 1 (Vg/ 2) ( 2ρ − ρ w )
ν′ = × = …(ii)
2l µ 2l µ

ν ′  2ρ − ρ w 
1/ 2
 2ρ − ρ w 
∴ =  or ν′ = ν  
ν  2ρ   2ρ 
1/ 2
 2ρ − 1
= 300   Hz Ans.
 2ρ 
∴ The correct option is (a).

INTRODUCTORY EXERCISE 18.3

1. Sound waves of frequency 660 Hz fall normally on a perfectly reflecting wall. The shortest
distance from the wall at which the air particles have maximum amplitude of vibration is …… m.
Speed of sound = 330 m/s. (JEE 1984)
2. If the frequencies of the second and fifth harmonics of a string differ by 54 Hz. What is the
fundamental frequency of the string?
3. A wire is attached to a pan of mass 200 g that contains a 2.0 kg mass as shown in the figure.
When plucked, the wire vibrates at a fundamental frequency of 220 Hz. An additional unknown
mass M is then added to the pan and a fundamental frequency of 260 Hz is detected. What is
the value of M?

Fig. 18.11

4. A wire fixed at both ends is 1.0 m long and has a mass of 36 g. One of its oscillation frequencies
is 250 Hz and the next higher one is 300 Hz.
(a) Which harmonics do these frequencies represent?
(b) What is the tension in the wire?
5. Two different stretched wires have same tension and mass per unit length. Fifth overtone
frequency of the first wire is equal to second harmonic frequency of the second wire. Find the
ratio of their lengths.
72 — Waves and Thermodynamics

T 2 2 T T

or  2  ω A i v1 =  2  ω 2 A r2 v1 +  2  ω 2 A t2 v 2
 v1   v1   v2 
A i2 A r2 A t2
or = + …(i)
v1 v1 v2
Further Ai + Ar = At …(ii)
Solving these two equations for A r and A t , we get
 v – v1 
Ar =  2  Ai
 v1 + v 2 
 2v 2 
and At =   Ai Hence proved.
 v1 + v 2 

INTRODUCTORY EXERCISE 18.4

1. Two pulses of identical shape overlap such that the displacement of the rope is momentarily
zero at all points, what happens to the energy at this time?
2. The pulse shown in figure has a speed of 10 cm/s.

Fig. 18.16
(a) If the linear mass density of the right string is 0.25 that of the left string, at what speed does
the transmitted pulse travel?
(b) Compare the heights of the transmitted pulse and the reflected pulse to that of the incident
pulse.
3. The harmonic wave y i = (2.0 × 10−3 )cos π(2.0x − 50t ) travels along a string toward a boundary
at x = 0 with a second string. The wave speed on the second string is 50 m/s. Write expressions
for reflected and transmitted waves. Assume SI units.
Exercises
LEVEL 1
Assertion and Reason
Directions : Choose the correct option.
(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true, but the Reason is false.
(d) If Assertion is false but the Reason is true.
1. Assertion : Two waves y1 = A sin (ωt + kx ) and y2 = A cos (ωt − kx ) are superimposed, then
x = 0 becomes a node.
Reason : At node net displacement due to two waves should be zero.
2. Assertion : Stationary waves are so called because particles are at rest in stationary waves.
Reason : They are formed by the superposition of two identical waves travelling in opposite
directions.
3. Assertion : When a wave travels from a denser medium to rarer medium, then amplitude of
oscillation increases.
Reason : In denser medium, speed of wave is less compared to a rarer medium.
4. Assertion : A wire is stretched and then fixed at two ends. It oscillates in its second overtone
mode. There are total four nodes and three antinodes.
3λ
Reason : In second overtone mode, length of wire should be l = , where λ is wavelength.
2
5. Assertion : If we see the oscillations of a stretched wire at higher overtone mode, frequency
of oscillation increases but wavelength decreases.
1
Reason : From v = f λ, λ ∝ as v = constant.
f
6. Assertion : Standing waves are formed when amplitudes of two constituent waves are
equal.
Reason : At any point net displacement at a given time is resultant of displacement of
constituent waves.
7. Assertion : In a standing wave x = 0 is a node. Then, total mechanical energy lying between
λ λ λ
x = 0 and x = is not equal to the energy lying between x = and x = .
8 8 4
Reason : In standing waves different particles oscillate with different amplitudes.
8. Assertion : Ratio of maximum intensity and minimum intensity in interference is 25 : 1.
The amplitude ratio of two waves should be 3 : 2.
2
I max  A1 + A2 
Reason : = 
I min  A1 − A2 
86 — Waves and Thermodynamics

9. Assertion : Three waves of equal amplitudes interfere at a point. Phase difference between
π
two successive waves is . Then, resultant intensity is same as the intensity due to individual
2
wave.
Reason : For interference to take place sources must be coherent.
10. Assertion : For two sources to be coherent phase difference between two waves at all points
should be same.
Reason : Two different light sources are never coherent.

Objective Questions
1. Two identical harmonic pulses travelling in opposite directions in a taut string approach each
other. At the instant when they completely overlap, the total energy of the string will be
A

(a) zero (b) partly kinetic and partly potential

(c) purely kinetic (d) purely potential
2. Three coherent waves having amplitudes 12 mm, 6 mm and 4 mm arrive at a given point with
successive phase difference of π/ 2. Then, the amplitude of the resultant wave is
(a) 7 mm (b) 10 mm
(c) 5 mm (d) 4.8 mm
3. Two transverse waves A and B superimpose to produce a node at x = 0.If the equation of wave A
is y = a cos ( kx + ωt ), then the equation of wave B is
(a) + a cos (kx − ωt ) (b) − a cos (kx + ωt )
(c) − a cos (kx − ωt ) (d) + a cos (ωt − kx)
4. In a stationary wave system, all the particles of the medium
(a) have zero displacement simultaneously at some instant
(b) have maximum displacement simultaneously at some instant
(c) are at rest simultaneously at some instant
(d) all of the above
5. In a standing wave on a string
(a) In one time period all the particles are simultaneously at rest twice
(b) All the particles must be at their positive extremes simultaneously once in a time period
(c) All the particles may be at their positive extremes simultaneously twice in a time period
(d) All the particles are never at rest simultaneously
6. If λ 1 , λ 2 and λ 3 are the wavelength of the waves giving resonance to the fundamental, first
and second overtone modes respectively in a string fixed at both ends. The ratio of the
wavelengths λ 1 : λ 2 : λ 3 is
(a) 1 : 2 : 3 (b) 1 : 3 : 5
1 1 1 1
(c) 1 : : (d) 1 : :
2 3 3 5
7. In a standing wave, node is a point of
(a) maximum strain (b) maximum pressure
(c) maximum density (d) All of these
 Chapter 18 Superposition of Waves — 87

8. For a certain stretched string, three consecutive resonance frequencies are observed as 105,
175 and 245 Hz respectively. Then, the fundamental frequency is
(a) 30 Hz (b) 45 Hz
(c) 35 Hz (d) None of these
9. A string 1 m long is drawn by a 300 Hz vibrator attached to its end. The string vibrates in three
segments. The speed of transverse waves in the string is equal to
(a) 100 m/s (b) 200 m/s
(c) 300 m/s (d) 400 m/s
10. Three resonant frequencies of string with both rigid ends are 90, 150 and 210 Hz. If the length
of the string is 80 cm, what is the speed of the transverse wave in the string?
(a) 45 m/s (b) 75 m/s
(c) 48 m/s (d) 80 m/s
11. The period of oscillations of a point is 0.04 s and the velocity of propagation of oscillation is
300 m/s. The difference of phases between the oscillations of two points at distance 10 m and
16 m respectively from the source of oscillations is
π
(a) 2π (b)
2
π
(c) (d) π
4
12. A transverse wave described by an equation y = 0.02 sin ( x + 30t ), where x and t are in metre
and second, is travelling along a wire of area of cross-section 1 mm 2 and density 8000 kgm −3 .
What is the tension in the string?
(a) 20 N (b) 7.2 N
(c) 30 N (d) 14.4 N

Subjective Questions
1. Two waves are travelling in the same direction along a stretched string. The waves are 90° out
of phase. Each wave has an amplitude of 4.0 cm. Find the amplitude of the resultant wave.
2. Two wires of different densities are soldered together end to end then stretched under tension
T . The wave speed in the first wire is twice that in the second wire.
(a) If the amplitude of incident wave is A, what are amplitudes of reflected and transmitted waves?
(b) Assuming no energy loss in the wire, find the fraction of the incident power that is reflected at
the junction and fraction of the same that is transmitted.
3. A wave is represented by
y1 = 10 cos ( 5x + 25 t )
where, x is measured in metres and t in seconds. A second wave for which
 π
y2 = 20 cos  5x + 25t + 
 3
interferes with the first wave. Deduce the amplitude and phase of the resultant wave.
4. Two waves passing through a region are represented by
y1 = (1.0 cm ) sin [( 3.14 cm −1 ) x − (157 s −1 ) t ]
and y2 = (1.5 cm ) sin [(1.57 cm −1 ) x − ( 314 s−1 ) t ]
Find the displacement of the particle at x = 4.5 cm at time t = 5.0 ms.
 Answers
Introductory Exercise 18.1
1. 49 : 1 2. (a) 4 : 1 (b) 16 : 1 3. 25 I0

Introductory Exercise 18.2

1. True 2. (a) 120 cm/s (b) 3 cm (c) zero
3. Yes, Yes 4. 15.7 m, 31.8 Hz, 500 m/s

Introductory Exercise 18.3

1. 0.125 2. 18 Hz 3. 0.873 kg
l
4. (a) 5th and 6th (b) 360 N 5. 1 = 3
l2

Introductory Exercise 18.4

Ar 1 At 4
1. Energy is in the form of kinetic energy 2. (a) 20 cms −1 (b) = , =
Ai 3 Ai 3
2 8
3. yr = × 10 −3 cos π (2.0x + 50t ), yt = × 10 −3 cos π (x − 50t )
3 3

Exercises
LEVEL 1
Assertion and Reason
1. (d) 2. (d) 3. (b) 4. (b) 5. (a) 6. (d) 7. (a) 8. (a) 9. (b) 10. (d)

Objective Questions
1. (c) 2. (b) 3. (c) 4. (d) 5. (a) 6. (c) 7. (d) 8. (c) 9. (b) 10. (c)
11. (d) 12. (b)

Subjective Questions
A 2 1 8 −1
1. 5.66 cm 2. (a) − , A (b) , 3. 26.46 cm, (5x + 25t + 0.714) rad 4. cm
3 3 9 9 2 2
5. (a) 0.02 s 6. (a) (i) (ii) (b) (i) (ii)

7. (a) 4.24 cm (b) 6.00 cm (c) 6.00 cm (d) 0.500 cm, 1.50 cm, 2.50 cm
8. (a) 2 cm (b) 2 π cm 9. 0.786 Hz, 1.57 Hz, 2.36 Hz, 3.14 Hz
10. (a) 163 N (b) 660 Hz 11. 19977 Hz 12. 2.142 m 13. 1.0 m
14. (a) 100 Hz (b) 700 Hz
15. One bridge at 6/11 m from one end and the other at 2/11 m from the other end.
16. The string should be pressed at 60 cm from one end.
 2 πx   2 πt  (b) 4.00 m/s
17. (a) y (x , t ) = (0.85 cm) sin   sin  (c) 0.688 cm
 0.3 cm   0.075 s 
18. (a) 3.0 m, 16.0 Hz (b) 1.0 m, 48.0 Hz (c) 0.75 m, 64.0 Hz
98 — Waves and Thermodynamics

19. (b) 2.80 cm (c) 277 cm (d) 185 cm, 7.9 Hz, 0.126 s, 1470 cm/s (e) 280 cm/s
(f) y (x , t ) = (5.6 cm) sin [ (0.0907 rad/ cm)x ] sin [(133 rad/ s)t ]
20. (a) 96.0 m/s (b) 461 N (c) 1.13 m/s, 426.4 m/s 2
π
21. (a) y = 0.28 sin  x − 3.0t +  (b) φ = ± 1.29 rad
 4
22. (i) Fundamental 12.5 Hz, third harmonic 37.5 Hz
πx 3 πx
(ii) (a) y = 0.1 sin sin 25 πt (b) y = 0.04 sin sin 75 πt
2 2
23. (a) 16 m, 5.33 m, 3.2 m (b) 6.25 Hz, 18.75 Hz, 31.25 Hz
24. (a) 8th and 9th (b) 2.16 m (c) 4.32 m
25. (a) y (x , t ) = 0.3 cos(2.0x + 40t + π )
(b) y (x , t ) = 0.3 cos(2.0x + 40t ) SI units
26. 8 s, inverted, 20 s upright
27. See the hints

LEVEL 2
Single Correct Option
1.(d) 2.(d) 3.(b) 4.(a) 5.(d) 6.(d) 7.(a) 8.(d) 9.(a) 10.(c)
11.(b) 12.(b) 13.(b) 14.(b) 15.(c)

More than One Correct Options

1.(a,c,d) 2.(a,d) 3.(a,b,d) 4.(a,c,d) 5.(a,c) 6.(b,d) 7.(b,d)

Comprehension Based Questions

1.(b) 2.(a) 3.(c)

Match the Columns

1. (a) → q (b) → r (c) → s (d) → r 2. (a) → r (b) → p (c) → r (d) → s
3. (a) → p (b) → p (c) → p (d) → r 4. (a) → s (b) → s (c) → p (d) → r
5. (a) → s (b) → p (c) → r (d) → s

Subjective Questions
7L µ1 πSρω 2 a2 π 2 a2T
1. (a) (b) No 2. 49% 3. 4.
2 F 4k 4l
5. (a) 0.14 s (b) –1.5 cm, 2.0 cm
f 3f 3 5 7 1 T
6. , f , … etc. when A is a node, f , f , f… etc. when A is an antinode. Here, f =
2 2 4 4 4 L µ

A, − tan−1 
5 3
7. 300 Hz 8. 0.05 sin πt − 0.0173 cos πt 9.
6  4
50 π 2ρω3 S
10. 40 kHz, 120 kHz 11. (a) ρω 2 S (b)
k k